\(\text{a) }A=x^2-4x+7\\ A=x^2-2.x.2+2^2+3\\ A=\left(x^2-2.x.2+2^2\right)+3\\ A=\left(x-2\right)^2+3\\ \text{Ta có : }\left(x-2\right)^2\ge0\\ \Rightarrow A=\left(x-2\right)^2+3\ge3\\ \text{Dấu }"="\text{xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Min\right)}=3\text{ xảy ra khi: }x=2\\ \)

\(\text{b) }B=x^2+8x\\ B=x^2+2\cdot x\cdot4+16-16\\ B=\left(x^2+2\cdot x\cdot4+4^2\right)-16\\ B=\left(x+4\right)^2-16\\ \text{Ta có : }\left(x+4\right)^2\ge0\\ \Rightarrow B=\left(x+4\right)^2-16\ge-16\\ \text{ Dấu }"="\text{ xảy ra khi: }\\ \left(x+4\right)^2=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\\ \text{ Vậy }B_{\left(Min\right)}=-16\text{ khi }x=-4\\ \)

\(\text{c) }C=2x^2+4x+15\\ C=\left(2x^2+4x+2\right)+13\\ C=2\left(x^2+2x+1\right)+13\\ C=2\left(x^2+2x+1^2\right)+13\\ C=2\left(x+1\right)^2+13\\ \text{Ta có : }\left(x+1\right)^2\ge0\\ \Rightarrow2\left(x+1\right)^2\ge0\\ \Rightarrow C=2\left(x+1\right)^2+13\ge13\\ \text{ Dấu }"="\text{ xảy ra khi: }\\ 2\left(x+1\right)^2=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\\ \text{Vậy }C_{\left(Min\right)}=13\text{ khi }x=-1\)

\(\text{d) }x^2+4y^2+z^2-2x-6z+8y+15\\ =x^2+4y^2+z^2-2x-6z+8y+1+4+9+1\\ =\left(x^2-2x+1\right)+\left[\left(2y\right)^2+8y+4\right]+\left(z^2-6z+9\right)+1\\ =\left(x^2-2x+1^2\right)+\left[\left(2y\right)^2+2\cdot2y\cdot2+2^2\right]+\left(z^2-2\cdot z\cdot3+3^2\right)+1\\ =\left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2+1\\ \text{Ta có : }\left(x-1\right)^2\ge0\forall x\\ \left(2y+2\right)^2\ge0\forall x\\ \left(z+3\right)^2\ge0\forall x\\ \left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2\ge0\forall x\\ \left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2+1\ge1\forall x\\ \)

Vậy biểu thức luôn nhận giá trị dương \(\forall x\)

Câu 3:

\(\text{a) }x^2+x+1\\ =x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{4}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ \text{Ta có : }\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ \text{ Vậy biểu thức luôn nhận giá trị dương}\text{ }\forall x\\ \)

\(\text{b) }2x^2+2x+1\\ =2x^2+2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{1}{2}\\ =2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\\ \text{Ta có: }2\left(x+\dfrac{1}{2}\right)^2\forall x\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\forall x\\ \text{Vậy giá trị của biểu thức luôn nhận giá trị dương }\forall x\\ \)

Câu 1:

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Thay \(a+b+c=0\) vào biểu thức ta được:

\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3=3abc\left(đpcm\right)\)

Vậy \(a^3+b^3=3abc\) khi \(a+b+c=0\)

\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\\ =x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+xyz+xyz\\ =\left(x^2y+x^2z+xyz+xz^2\right)+\left(xy^2+y^2z+xyz+yz^2\right)\\ =x\left(xy+xz+yz+z^2\right)+y\left(xy+yz+xz+z^2\right)\\ =\left(x+y\right)\left(xy+yz+xz+z^2\right)\\=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\\=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\\ =\left(x+y\right)\left(y+z\right)\left(x+z\right) \)

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